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प्रश्न
If 3 + k, 18 – k, 5k + 1 are in A.P. then find k
योग
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उत्तर
3 + k, 18 – k, 5k + 1 are in A.P
⇒ 2b = a + c if a, b, c are in A.P
∴ `underbrace(3 + "k")_"a", underbrace(18 - "k")_"b", underbrace(5"k" + 1)_"c"`
2b = a + c
⇒ 2(18 – k) = 3 + k + 5k + 1
36 – 2k = 4 + 6k
6k + 2k = 36 – 4
8k = 32
k = `32/8` = 4
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